csvdc(3P)

NAME

csvdc - compute the singular value decomposition of a general matrix A.

SYNOPSIS

CALL DSVDC (DA, LDA, N, P, DSVALS, DE, DLSVEC, LDL, DRSVEC, LDR, DWORK, JOB, INFO)

CALL SSVDC (SA, LDA, N, P, SSVALS, SE, SLSVEC, LDL, SRSVEC, LDR, SWORK, JOB, INFO)

CALL ZSVDC (ZA, LDA, N, P, ZSVALS, ZE, ZLSVEC, LDL, ZRSVEC, LDR, ZWORK, JOB, INFO)

CALL CSVDC (CA, LDA, N, P, CSVALS, CE, CLSVEC, LDL, CRSVEC, LDR, CWORK, JOB, INFO)

void dsvdc(double ∗dx, long int ldx, long int n, long int p,
double ∗s, double ∗e, double ∗du, long int ldu, double ∗v, long int ldv, long int job, long int ∗info)

void ssvdc(float ∗sx, long int ldx, long int n, long int p, float
∗s, float ∗e, float ∗su, long int ldu, float ∗v, long int ldv, long int job, long int ∗info)

void zsvdc(doublecomplex ∗zx, long int ldx, long int n,
long int p, doublecomplex ∗s, doublecomplex ∗e, doublecomplex ∗u, long int ldu, doublecomplex ∗v, long int ldv, long int job, long int ∗info)

void csvdc(complex ∗cx, long int ldx, long int n,
long int p, complex ∗s, complex ∗e, complex ∗cu, long int ldu, complex ∗v, long int ldv, long int job, long int ∗info)

ARGUMENTS

xAMatrix A.

LDALeading dimension of the array A as specified in a dimension or
type statement. LDA >= max(1,N).

NNumber of rows of the matrix A. N >= 0.

PNumber of columns of the matrix A. P >= 0.

xSVALSOn exit, the singular values of A arranged in descending
order of magnitude.

xEOn exit, normally contains zero, but see INFO below.

xLSVECOn exit, the matrix of left singular vectors; not referenced if
the a digit of JOB = 0.

LDLLeading dimension of LSVEC as specified in a dimension or type statement. LDL >= max(1,N).

xRSVECOn exit, the matrix of right singular vectors; not referenced if
the b digit of JOB = 0.

LDRLeading dimension of RSVEC as specified in a dimension or
type statement. LDR >= max(1,P).

xWORKScratch array with a dimension of N.

JOBInteger in the form ab; determines operation subroutine will perform:
a = 0 do not compute the left singular vectors
a = 1 return the n left singular vectors in LSVEC
a ∗ 2 return the first min(N,P) singular vectors in LSVEC
b = 0 do not compute the right singular vectors
b = 1 return the right singular vectors in RSVEC

INFOOn exit, the singular values and their corresponding singular
vectors SVALS(INFO+1), SVALS(INFO+2),...,SVALS(min(N,P)) are correct. If INFO = 0 then all singular values and singular vectors are correct. The matrix B defined as LSVECT ∗ A ∗ RSVEC is the bidiagonal matrix with S on its diagonal and E on its superdiagonal. Therefore the singular values of A and B are the same.

SAMPLE PROGRAM

      PROGRAM TEST
      IMPLICIT NONE
C
      INTEGER           JOB, LDL, LDR, LDX, N, P
      PARAMETER        (JOB = 21)
      PARAMETER        (N = 3)
      PARAMETER        (P = 3)
      PARAMETER        (LDL = N)
      PARAMETER        (LDR = P)
      PARAMETER        (LDX = N)
C
      DOUBLE PRECISION EPSLON, EXCEPT(P), LSVALS(LDL,N)
      DOUBLE PRECISION RSVALS(LDR,P), SVALS(N), WORK(N), X(LDX,P)
      INTEGER           I, ICOL, INFO, IRANK, IROW
C
      EXTERNAL          DSVDC
      INTRINSIC         ABS, SQRT
C
C     Initialize the array X to store the matrix X shown below.
C
C         1 1 3
C     X = 0 1 1
C         1 0 1
C
      DATA X / 1.0D0, 0.0D0, 1.0D0, 1.0D0, 1.0D0, 0.0D0,
     $         3.0D0, 1.0D0, 1.0D0 /
C
      PRINT 1000
      PRINT 1010, ((X(IROW,ICOL), ICOL = 1, N), IROW = 1, N)
      CALL DSVDC (X, LDX, N, P, SVALS, EXCEPT, LSVALS, LDL,
     $            RSVALS, LDR, WORK, JOB, INFO)
      PRINT 1020
      PRINT 1030, SVALS
C
C     Compute the unit roundoff
C
      EPSLON = 1.0D0
   10 IF (DBLE (1.0D0 + EPSLON) .NE. 1.0D0) THEN
        EPSLON = EPSLON / 2.0D0
        GO TO 10
      END IF
C
C     Make a conservative estimate of the rank of A.
C
      IRANK = 0
      EPSLON = SQRT (EPSLON + EPSLON)
      DO 20, I = 1, N
        IF (ABS(SVALS(I)) .GT. EPSLON) THEN
          IRANK = IRANK + 1
        END IF
   20 CONTINUE
      PRINT 1040, IRANK
C
1000 FORMAT (1X, ’X:’)
1010 FORMAT (3(3X, F4.1))
1020 FORMAT (/1X, ’Singular values of X:’)
1030 FORMAT (3X, F4.1)
1040 FORMAT (/1X, ’The rank of X is ’, I1)
C
      END

SAMPLE OUTPUT

X:
    1.0    1.0    3.0
    0.0    1.0    1.0
    1.0    0.0    1.0

Singular values of X:
    3.7
    1.0
    0.3

The rank of X is 3